YUFRIDIN3
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The Common The Common - - Collector Amplifier Collector Amplifier The common-collector amplifier is usually referred to as the emitter follower because there is no phase inversion or voltage gain. The output is taken from the emitter. The common-collector amplifier’s main advantages are it’s high current gain and high input resistance. The Common The Common - - Collector Amplifier Collector Amplifier The input resistance can be determined by the simplified formula below. R in(base) ≅β ac (r’ e + R e ) R in(tot) =R in(base) //R 1 // R 2 Output Resistance is very low: R out ≅ R E (R s /β ac ) The Common The Common - - Collector Amplifier Collector Amplifier The Voltage Gain can be determined by the simplified formula below. A v = R e /(r’ e + R e ) and is always less than 1 but if r’ e << R e then A v ≅ 1 is a good approximation. The current Gain: A i ≅β ac with R 1 // R 2 >> R e β ac The Power Gain: A p = A v A i ≅ A i with A v ≅ 1 Example Example
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Transcript of YUFRIDIN3
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The CommonThe Common--Collector AmplifierCollector AmplifierThe common-collector amplifier is usually referred to as the emitter follower because there is no phase inversion or voltage gain. The output is taken from the emitter. The common-collector amplifiers main advantages are its high current gain and high input resistance.
The CommonThe Common--Collector AmplifierCollector Amplifier
The input resistance can be determined by the simplified formula below.
Rin(base) ac(re + Re)Rin(tot) = Rin(base)//R1 // R2
Output Resistance is very low: Rout RE (Rs /ac)
The CommonThe Common--Collector AmplifierCollector Amplifier
The Voltage Gain can be determined by the simplified formula below. Av = Re/(re + Re) and is always less than 1 but if r e > Reac
The Power Gain: Ap = AvAi Ai with Av 1
Example Example
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ExampleExampleExample 6.9: Calculate Input Resistance, Voltage Gain, Current Gain and Power Gain for the given circuit.
DC analysis: VE = R2 /(R1+R2 ) - VBE = 4.3V
DC analysis: IE = VE /RE = 4.3mA
AC Analysis: re =25mV/IE = 5.8
ExampleExampleExample 6.9: Calculate Input Resistance, Voltage Gain, Current Gain and Power Gain for the given circuit.
Re = RE // RL : 1k//1k = 500
Rin(tot) = Rin(base)//R1 // R2 = 8.16k
Rin(base) ac(re + Re) but re is just 5.8 so Rin(base) ac(Re) = 175(500) = 87.5k
ExampleExampleExample 6.9: Calculate Input Resistance, Voltage Gain, Current Gain and Power Gain for the given circuit.
= 2mA/123A = 16.3
Av = Re/(re + Re) = 500/505.8 = 0.989
Ai = Ie/Iin = (Ve/Re)/(Vin/Rin(tot))=
= (VbAv/500)/(1/8.16k)
ConclusionConclusion
The output resistance is very low. This makes it useful for driving low impedance loads.
The current gain(Ai) is approximately ac.
The power gain is approximately equal to the current gain(Ai).
The voltage gain is approximately 1.
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The CommonThe Common--Collector AmplifierCollector AmplifierThe darlington pair is used to boost the input impedance to reduce loading of high output impedance circuits. The collectors are joined together and the emitter of the input transistor is connected to the base of the output transistor. The input impedance can be determined the formula below.
Rin = ac1ac2Re
The CommonThe Common--Collector AmplifierCollector Amplifier
Tutorial and AssignmentTutorial and Assignment
Solve Example 6.9 and 6.10 and send to me on Wednesday before class.
Tutorial II : Qs 18 to 23 pages 319-320. To be discussed on Thursday.
